Question

If \(\mathrm{rms}\) speed of a gas is \(\mathrm{v}_{\mathrm{rms}}=1840 \mathrm{~m} / \mathrm{s}\) and its density \(\rho=8.99 \times 10^{-2} \mathrm{~kg} / \mathrm{m}^{3}\), the pressure of the gas will be (A) \(1.01 \times 10^{3} \mathrm{Nm}^{-2}\) (B) \(1.01 \times 10^{5} \mathrm{Nm}^{-2}\) (C) \(1.01 \times 10^{7} \mathrm{Nm}^{-2}\) (D) \(1.01 \mathrm{Nm}^{-2}\)

Short answer

The pressure of the gas is calculated using the equation \(P = \rho \cdot \frac{v_{\text{rms}}^2}{3}\), where \(\rho\) is the density and \(v_{\text{rms}}\) is the rms speed. Substituting the given values, we get \(P = (8.99 \times 10^{-2} \mathrm{~kg}\, / \mathrm{m}^{3}) \cdot \frac{(1840\mathrm{~m}\, / \mathrm{s})^2}{3} = 1.01 \times 10^{5} \mathrm{Nm}^{-2}\). Therefore, the correct answer is (B) \(1.01 \times 10^{5} \mathrm{Nm}^{-2}\).

Step-by-step solution

Calculate the number of moles in the gas

To calculate the number of moles in the gas, we need to use the following relationship between rms speed (v_rms), molar mass (M), and density: \[ \rho = \frac{nM}{V}. \] As the density is given, we can calculate the number of moles \(n\) per unit volume \(V\). Rearrange the equation for \(n\): \[ n = \frac{\rho V}{M}. \] We need to find the molar mass (M) next. To find the molar mass, we will use the rms speed and the relationship: \[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}}, \] where \(R\) is the gas constant and \(T\) is the temperature. Rearrange the equation for molar mass: \[ M = \frac{3RT}{v_{\text{rms}}^2}. \]

Calculate the pressure

Now, we will use the ideal gas equation to find the pressure: \[ PV = nRT. \] Plug the expression for \(n\) in terms of \(\rho\), \(V\), and \(M\): \[ P = \frac{nRT}{V} = \frac{\rho R T}{M}. \] We know the value of \(\rho\), \(v_{\text{rms}}\), and will replace \(M\) with the expression from step 1: \[ P = \frac{\rho R T}{\frac{3RT}{v_{\text{rms}}^2}} = \rho \cdot \frac{v_{\text{rms}}^2}{3}. \] Now, we can calculate the pressure using the rms speed and density: \[ P = (8.99 \times 10^{-2} \mathrm{~kg}\, / \mathrm{m}^{3}) \cdot \frac{(1840\mathrm{~m}\, / \mathrm{s})^2}{3} = 1.01 \times 10^{5} \mathrm{Nm}^{-2}. \] So, the correct answer is (B) \(1.01 \times 10^{5} \mathrm{Nm}^{-2}\).

Key concepts

Understanding rms Speed of Gas

The root mean square (rms) speed of a gas is a measure used to describe the speed of particles in a gas sample. It is derived from the kinetic theory of gases, which assumes that gas particles move randomly and collide with each other and the walls of their container.
The rms speed gives an average velocity accounting for the square root mean of all particle speeds.To calculate the rms speed (\(v_{\text{rms}}\)), we use the formula:\[v_{\text{rms}} = \sqrt{\frac{3RT}{M}},\]where \(R\) is the universal gas constant, \(T\) represents the absolute temperature, and \(M\) is the molar mass of the gas. This formula is derived from the principles that link energy to temperature.
The rms speed increases with temperature, since more heat energy is available to speed up gas particles. Understanding what rms speed represents helps explain how gases behave under different temperatures and conditions.

Pressure Calculation Using Kinetic Theory

Calculating the pressure exerted by a gas is key in understanding how gases interact with their environment. According to the kinetic theory, pressure (\(P\)) is the result of collisions between gas particles and the walls of their container. These collisions exert a force over a given area.The pressure calculation combines concepts of particle dynamics and gas properties, such as density and speed. By rearranging the ideal gas law and incorporating rms speed and gas density, we derive the expression:\[P = \rho \cdot \frac{v_{\text{rms}}^2}{3},\]where \(\rho\) is the gas density and \(v_{\text{rms}}\) is the root mean square speed. This shows that pressure is directly proportional to both the density and the square of the rms speed of the gas particles.
In practical terms, an increase in either the number of particles or their speed will lead to a higher pressure, explaining the behavior of gases under varying conditions.

The Ideal Gas Law

The ideal gas law is a fundamental equation that relates key properties of gases: pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and the number of moles (\(n\)) of the gas. Expressed as:\[PV = nRT,\]where \(R\) is the ideal (universal) gas constant, this equation provides a good approximation for the behavior of many gases under a variety of conditions.
The ideal gas law assumes that the interaction between gas particles is negligible, and their movements are random and independent (except during collisions). This assumption generally holds true except under high pressure or low temperature, where deviations from ideal behavior occur.In practical terms, this law allows you to solve for any single variable if the others are known. By understanding the interplay of these properties, one can predict how a gas will respond to changes in external conditions, such as temperature increases or volume restrictions.
This predictive capability is crucial in fields ranging from chemistry and physics to engineering and meteorology.